How to evaluate $\int_0^a\frac{x^4}{(a^2+x^2)^4}dx$? I have to evaluate 

$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,{\rm d}x$$

I tried to substitute $x=a\tan\theta$ which then simplifies to
$$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta\, {\rm d}\theta$$.
Now,its quite hectic to solve this. Is there any other method out?
 A: It's not becoming "hectic" at all:
$$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta=$$
$$\frac1{a^5}\int_0^\frac\pi4(\sin\theta\cos\theta)^2\sin^2\theta d\theta=$$
$$\frac1{8a^5}\int_0^\frac\pi4(\sin^22\theta)(1-\cos2\theta) d\theta=$$
$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta)(1-\cos2\theta) d\theta=$$
$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\cos2\theta+\cos4\theta\cos2\theta) d\theta=$$
$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\cos2\theta+\frac12(\cos6\theta+\cos2\theta))d\theta=$$
$$\frac1{16a^5}\int_0^\frac\pi4(1-\cos4\theta-\frac12\cos2\theta+\frac12\cos6\theta)d\theta=$$
$$\frac1{16a^5}(\theta-\frac14\sin4\theta-\frac14\sin2\theta+\frac1{12}\sin6\theta)|_{0}^{\pi/4}=$$
$$\frac1{16a^5}(\frac\pi4-\frac13)=\frac1{192}(3\pi-4)$$
...and Mathematica gives the same result.
A: HINT:
If you want to solve $\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta$ then
$1)$ Put $\theta=8t$ and then $\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta d\theta=\frac1{8a^5}\int_0^{2\pi}\sin^4{8t}\cos^2{8t} dt$
$2)$ Put $z=e^{it}$ and after some substitutions you will have an integral of a rational function $\frac{P(z)}{Q(z)}$

Now you  can find the poles of the function in the region surrounded by the unit circle and apply the residue theorem.

With the above substitution: $$\sin{t}=\frac{z-\frac{1}{z}}{2i}$$ $$\cos{t}=\frac{z+\frac{1}{z}}{2}$$
You can also find similar relations for $$\sin{8t},\cos{8t}$$
A: Once it is clear that the original integral is just $\frac{C}{a^5}$, with
$$ C=\int_{0}^{\pi/4}\sin^4(\theta)\cos^2(\theta)\,d\theta\stackrel{\theta\mapsto\varphi/2}{=}\frac{1}{2}\int_{0}^{\pi/2}\left(\frac{1-\cos\varphi}{2}\right)^2\left(\frac{1+\cos\varphi}{2}\right)\,d\varphi $$
one may simply expand everything and exploit
$$ \int_{0}^{\pi/2}\cos^{2m}(\varphi)\,d\varphi = \frac{\pi}{2\cdot 4^m}\binom{2m}{m},\qquad \int_{0}^{\pi/2}\cos^{2m+1}(\varphi)\,d\varphi = \frac{4^m}{(2m+1)\binom{2m}{m}} $$
which are crucial in many proofs of Wallis' product or Stirling's inequality, for instance.
Maybe not the most efficient approach, but for sure it is straightforward. It leads to $C=\frac{3\pi-4}{192}$.
A: This is a rather standard rational function integral, and we only have to decompose it into partial fractions.
To have lighter calculations, we'll set $T=a^2+x^2$. Then
$$x^4=T^2-2a^2x^2-a^4=T^2-2a^2T+a^4$$
which yields the decomposition
\begin{align}
\frac{x^4}{(a^2+x^2)^4}&=\frac{T^2-2a^2T+a^4}{T^4}=\frac1{T^2}-\frac{2a^2}{T^3}+\frac{a^4}{T^4}\\
&=\frac1{(a^2+x^2)^2}-\frac{2a^2}{(a^2+x^2)^3}+\frac{a^4}{(a^2+x^2)^4}.
\end{align}
Now the integrals $\;I_n=\int\frac{\mathrm d x}{(a^2+x^2)^n}$ can be computed recursively, from
\begin{cases}
I_1=\dfrac1a\arctan \dfrac xa, \\
I_{n+1}=\dfrac1{2na^2}\dfrac x{(a^2+x^2)^n}+\dfrac{2n-1}{2na^2} I_n.
\end{cases}
The recurrence relation can be obtained with an integration by parts of $I_n$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation}
\left.\int_{0}^{a}{x^{4} \over \pars{a^{2} + x^{2}}^4}\,\dd x
\,\right\vert_{\ a\ \not=\ 0} =
{1 \over a^{3}}\int_{0}^{1}{x^{4} \over \pars{1 + x^{2}}^4}
\,\dd x
\label{1}\tag{1}
\end{equation}

\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{x^{4} \over \pars{1 + x^{2}}^4}\,\dd x}
=
\left. -\,{1 \over 6}\,\totald[3]{}{b}\int_{0}^{1}{x^{4} \over b + x^{2}}\,\dd x\,\right\vert_{\ b\ =\ 1}
\\[5mm] = &\
\left. -\,{1 \over 6}\,\totald[3]{}{b}\int_{0}^{1}
\pars{-b + x^{2} + {b^{2} \over b + x^{2}}}\,\dd x\,\right\vert_{\ b\ =\ 1}
\\[5mm] = &\
-\,{1 \over 6}\,\totald[3]{}{b}\bracks{b^{3/2}\int_{0}^{1}
{1 \over \pars{x/\root{b}}^{2} + 1}\,{\dd x \over \root{b}}}_{\ b\ =\ 1}
\\[5mm] = &\
-\,{1 \over 6}\,\totald[3]{}{b}\pars{b^{3/2}\int_{0}^{1/\root{b}}
{\dd x \over x^{2} + 1}}_{\ b\ =\ 1} =
\left. -\,{1 \over 6}\,\totald[3]{\bracks{b^{3/2}\arctan\pars{b^{-1/2}}}}{b}\right\vert_{\ b\ =\ 1}
\\[5mm] = &\
\left.{\pars{b + 3}\pars{3b - 1} \over 8b\pars{b + 1}^{3}} -
{3\,\mrm{arccot}\pars{\root{b}} \over 8b^{3/2}}
\,\right\vert_{\ b\ =\ 1} = \bbx{3\pi - 4 \over 192} \approx
0.0283
\end{align}
$$
\bbx{\left.\int_{0}^{a}{x^{4} \over \pars{a^{2} + x^{2}}^4}\,\dd x
\,\right\vert_{\ a\ \not=\ 0} =
{3\pi - 4 \over 192a^{3}}}
$$
A: The change of variable $x=a\tan\theta$ give us
$$I=\int_0^a\frac{x^4}{(a^2+x^2)^4}\, dx=\frac{1}{a^3}\int_0^{\pi/4}\sin^4(\theta)\cos^2(\theta) \,d\theta.$$
Now write $\cos^2(\theta)=1-\sin^2(\theta)$, such that
$$I=\frac{1}{a^3}\int_0^{\pi/4}\left[\sin^4(\theta)-\sin^6(\theta)\right] \,d\theta.$$
Use then the reduction formula:
$$\int \sin^m(x)\,dx=-\frac{\cos(x)\sin^{m-1}(x)}{m}+\frac{m-1}{m}\int\sin^{-2+m}(x)\,dx.$$
Using this, one gets

$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,dx =\frac{3\pi-4}{192a^3}$$

A: Hint:
Use these identities
$$\sin^2t=\dfrac{1-\cos2t}{2}~~~~~,~~~~~~\cos^2t=\dfrac{1+\cos2t}{2}$$
